Question: The scalar field $f(x, y) = 2x^2 - y^2 + 4x + 2y - 1$ has a critical point at $(-1, 1)$. How does the second partial derivative test classify this critical point? Choose 1 answer: Choose 1 answer: (Choice A) A Local maximum (Choice B) B Local minimum (Choice C) C Saddle point (Choice D) D The test is inconclusive
The second partial derivative test uses the quantity below, evaluated at the critical point we wish to classify. $H = f_{xx}f_{yy} - f_{xy}f_{yx}$ $H < 0$ implies a saddle point. $H > 0$ and $f_{xx} > 0$ implies a local minimum. $H > 0$ and $f_{xx} < 0$ implies a local minimum. $H = 0$ means the test is inconclusive. Let's calculate $H$. First we need all the regular partial derivatives. $\begin{aligned} f_x &= 4x + 4 \\ \\ f_y &= -2y + 2 \end{aligned}$ Now we can find all the second order partial derivatives. $\begin{aligned} f_{xx} &= 4 \\ \\ f_{yx} &= 0 \\ \\ f_{xy} &= 0 \\ \\ f_{yy} &= -2 \end{aligned}$ Therefore, $H = (4)(-2) - (0)(0) = -8$. Because $H$ is negative, we know that the critical point is a saddle point.